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Implementing binary search in JavaScript to return the index if the searched number exist
We are required to write a JavaScript function that takes in a sorted array of numbers as the first argument and a search number as the second argument.
If the search number exists in the array, we need to return its index in the array, otherwise we need to return -1.
We have to do this making use of the binary search algorithm. The binary search algorithm is basically a divide and conquer algorithm which recursive divides the array into halves until it converses to a singleton element.
The sorting of array is necessary of binary search algorithm in this case, as it makes deciding which part to divide easy for us.
Example
const arr = [-3, -1, 4, 7, 9, 11, 14, 22, 26, 28, 36, 45, 67, 78, 88, 99];
const binarySearch = (arr = [], num) => {
let l = 0;
let r = arr.length - 1;
while(l <= r){
const mid = Math.floor((l + r) / 2); if(num == arr[mid]){
return mid;
}
else if(num < arr[mid]){
r = mid - 1;
}
else{
l = mid + 1;
};
};
return -1
};
console.log(binarySearch(arr, 22));
console.log(binarySearch(arr, 56));
console.log(binarySearch(arr, 11));Output
And the output in the console will be −
7 -1 5
Updated on: 21-Nov-2020
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